Answer:
Average speed = 3.63 m/s
Explanation:
The average speed during any time interval is equal to the total distance travelled divided by the total time.
That is,
Average speed = distance/ time
Let d represent the distance between A and B.
Let t1 be the time for which she has the higher speed of 5.15 m/s. Therefore,
5.15 = d/t1.
Make d the subject of formula
d = 5.15t1
Let t2 represent the longer time for the return trip at 2.80 m/s . That is,
2.80 = d/t2.
Then the times are t1 = d/5.15 5 and
t2 = d/2.80.
The average speed vavg is given by the following equation.
avg speed = Total distance/Total time
Avg speed = d + d/t1 + t2
Where
Total distance = 2d
Total time = t1 + t2
Total time = d/5.15 + d/2.80
Total time = (2.8d + 5.15d)/14.42
Total time = 7.95d/14.42
Total time = 0.55d
Substitute total distance and time into the formula above.
Avg speed = 2d / 0.55d
Avg Speed = 3.63 m/s
In this problem we have the electric field intensity E:
E = 6.5 × newtons/coulomb
We have the magnitude of the load:
q = 6.4 × coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × )(6.5 × )(1.2 × )
PE = 5.0 x joules
None of the options shown is correct.
Answer:
Option B) This minimizes the harmful side effects of the radiations
Explanation:
Half-life is the time taken for the decay of an radio-active atom in which it disintegrates such that it becomes half of its value at the beginning.... The nuclei should be in active mode for a longer duration sufficient for the treatment of the condition but these nuclei should have a sufficient shorter half life so that they don't get enough time to cause any damage to the health of the person other than treating the cause.
A shorter half life gives the assurance that the radiation after the treatment will leave the body without getting accumulated and cause harm to the body cells and other organs.